mathschallenge.net

Problem

If we consider 2-digit numbers, there are exactly 18 numbers that contain the digit one:

10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
21, 31, 41, 51, 61, 71, 81, 91

As there are ninety 2-digit numbers, the probability of a 2-digit number containing a one is 18/90 = 1/5.

What proportion of 3-digit numbers contain the digit one?

Solution

There are three mutually exclusive cases for consideration:
1**, (1')1*, and (1')(1')1.

First Digit	Second Digit	Third Digit	Combinations
1	0-9	0-9	11010 = 100
2-9	1	0-9	8110 = 80
2-9	2-9 + 0	1	891 = 72
Total			252

There are 91010 = 900 numbers that contain three digits.

Hence the proportion of 3-digit numbers containinig the digit one is 252/900 = 7/25.

Would you expect the proportion of 4-digit numbers containing the digit one to be more or less?
What about n-digit numbers?